Assuming the Earth's mass is evenly distributed around the centre, it's pretty intuitive that there as a net effect of zero gravity at the centre, as there's an equal force being applied in every direction, each cancelling out the effect of the one in the opposite direction.

Basically in electrodynamics it turns out that the divergence of the electric field is zero, and because magnetic fields do not have monopoles like electric fields do shows that magnetism and gravity are not similar, by virtue of the fact that gravity is monopolar. Basically the field equations that describe this subject can be very difficult to explain in terms of basic math, it requires vector calculus, plus I don't really know for a certain if there is a definite answer to your question in terms of explaining things apart from the foundations that describe the fields you want to talk about. Basically the math describes the subject and even though there has been no experiment to prove that gravity doesn't exist at the center of the earth, the equations of the fields I described "imply" such a condition! Physics as with any science is always extended by experiments as much as people can do them and if the equation works according to theory then its a good nights rest.

The problem I see here is you're not accounting for the direction of each particle.

Says who? r in the x-direction, r in the y-direction, and r in the z-direction yield forces in each of those directions.

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It's also a convenient approximation the equation makes.

Where?? The equation says Mass at distance "r" away from you has this amount of force. When you are in the center then mass is located where? The entire mass of the earth is not located in a single point in the center.

I think you all misunderstood. My roommate and I are in perfect agreement that there is no net gravitational pull from Earth at the center of the Earth, the argument is over when humans first figured that out, before or after 1864.

I stated that newton had it all figured out when he came up with GMm/r2 in the 1600s. Makes sense doesn't it? He probably studied different shapes and such.. and he did invent co-discover calculus.

if you cut the sphere up into circular slices than you can realistically calculate it.

Google for "washer method" +calculus

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Speaking as someone who has recently been beside the sea, I do not believe that the world's mass is evenly distributed around the centre you'd attribute it when modelling as a sphere. Having read about geology, I know that mass is not evenly distributed according to strata. Having read about space, I know that the Earth is not an inertial frame. Furthermore, since the stuff in the middle of the Earth is fluid, the Earth doesn't even have a constant centre of gravity from which you can disregard most of the above.

Furthermore, since the stuff in the middle of the Earth is fluid, the Earth doesn't even have a constant centre of gravity from which you can disregard most of the above.

you're right! its just that the mass of the earth is megagigatons and your version of earths cog would probably have an uncertainty on the scale of tiny lengths, I can't quote any source however but it would be small.

its just that the mass of the earth is megagigatons and your version of earths cog would probably have an uncertainty on the scale of tiny lengths, I can't quote any source however but it would be small.

The tiny lengths you speak of are of at least the same order of magnitude as me. Ignoring that the original question is stupid even if you're naive ("what if you were at the centre of the Earth, but, ummm, somehow not affected by the flow or the pressure of the fluid down there, but the fluid is still down there for the purposes of gravity?"), there's the fact that the Earth and moon actually orbit each other, around their common centre of mass. So the answer is: if you were momentarily at the centre of mass of the Earth but otherwise unaffected by a large part of what dictates the centre of mass of the Earth, you'd be pulled towards the moon.

We know today that at the Earth's center of gravity the Earth's gravity would have no net effect on an object.

"I've decided what conclusion I want to reach"

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However, me and my roommate are disagreeing about how long this has been known.

"We've decided that everybody else must agree with us"

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The question in debate is whether or not it was scientifically accepted to say that there would be gravity at the center of the Earth at the time that Jules Verne wrote "Journey to the Centre of the Earth".

"We can't see how, were one large body having no gravitational effect on us, any other body could either"

So the answer is: if you were momentarily at the centre of mass of the Earth but otherwise unaffected by a large part of what dictates the centre of mass of the Earth, you'd be pulled towards the moon.

I guess you didn't understand that the net force it zero when R=0 in the original equation. And I don't know if you've worked with uncertainties before but those comments are relatively small of which you speak and that is exactly where the uncertainty comes to play, whether the fluid is there or somewhere else, the scale of the earth dictates that it has a cog, which can possibly change because of fluid dynamics and this uncertainty is NOT ignored, its a part of every calculation that is made today in experimental physics. There is uncertainty in anything measured and there is a method that explains that. Obviously you don't read into science journals, there everything has uncertainty to some degree. I guess you could say science is naive for not propagating uncertainties but thats not the case.

I guess you didn't understand that the net force it zero when R=0 in the original equation.

There is no R in "the original equation". Assuming you mean r, which would be a surprising mistake from someone that is highly scientific, net force is undefined in the original equation. At best you mean that the original equation applies only where gravity is modelled as emanating from a point mass?

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I don't know if you've worked with uncertainties before but those comments are relatively small

One of the comments I made was that the moon and the Earth together make the major local gravitational field, and that the two orbit each other. Having looked it up, the centre of mass of the total system is about 4,700km from the centre of the Earth (3/4 of the Earth's radius) and isn't static because the Earth and the moon are constantly rotating.

Are you conveniently ignoring that, or are you arguing that 3/4 of the Earth's radius is "relatively small" when we're discussing an object the size of the Earth?

Gravity map of the Earth, courtesy of NASA:{"name":"PIA12146.gif","src":"\/\/djungxnpq2nug.cloudfront.net\/image\/cache\/3\/5\/357acab53f3b700d22b4b1c993f38486.gif","w":256,"h":256,"tn":"\/\/djungxnpq2nug.cloudfront.net\/image\/cache\/3\/5\/357acab53f3b700d22b4b1c993f38486"}(http://photojournal.jpl.nasa.gov/catalog/PIA12146 in case the gif doesn't animate)

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You are wrong, Thomas Harte. The only time you can make an approximation that "centre-of-mass = point source of gravity" is when you are dealing with a single spherical shell (or really really far away from the component bodies, at a distance of a light year, you could probably take the center of mass of the entire solar system and treat it as a point source).

Or do you really think that during the high tide the gravity is 9 times stronger on the side of the earth that is closest to the moon compared to the side of the earth farthest from the moon? Plug in your 'centre of mass=point source' idea into the equation and you'll see that people must be able to jump quite a bit higher when there's high tide.

You are wrong, Thomas Harte. The only time you can make an approximation that "centre-of-mass = point source of gravity" is when you are dealing with a single spherical shell

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As verthex said, the effects you are clamouring are negligible.

The effect of a body of mass 7.36 × 1022 kilograms that is just 384,403 km away is negligible? That it's sufficient to move the centre of gravity of the two bodies combined by about 4783.5 km, but, you know, negligible on a much, much smaller scale?

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Or do you really think that during the high tide the gravity is 9 times stronger on the side of the earth that is closest to the moon compared to the side of the earth farthest from the moon?

You obviously fail to understand what I'm saying. The thrust of my argument is twofold: (i) the centre of gravity of the Earth is quite variable compared to the size of a human being; (ii) even if the human being were positioned such that the gravitational effect of the Earth were zero, there's another very large body very nearby that would prevent apparent weightlessness.

You seem to be saying that the second isn't true because the pull of the Earth over objects situated on the Earth is greater than the pull of the moon, which simply isn't relevant.

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NASA said:

Gravity is determined by mass. Earth’s mass is not distributed equally, and it also changes over time.

Mentioned on gnolam's link. NASA seem to think that the change in distribution of mass is sufficient to need to be mentioned in the brief text following a diagram about how mass is not distributed evenly across the globe.

Ok everyone let me honestly say that I dont try to disect every tiny argument word for word because THIS is not a scientific discussion. I've sat around with people that do this in a room all day and they don't do anything but pull up facts and figures and make me go crazy!

This suggests you really don't understand the problem with your tide argument. The Earth has a radius of about 6,371km. Of that, 2,260km is fluid, which extends continuously from about 1,220km from the centre. (EDIT3: 15% of the mass of the Earth is fluid, it seems)

If you want to plug numbers in, work out the effect of the moon on that, then the effect of that on the person in the centre.

EDIT: to sum up again, since you seem to be cutting and chopping my argument according to whim; the centre of mass of Earth is not fixed because a large part of the planet is fluid and there's a really big body nearby that affects the fluids on this planet. You would therefore not experience weightlessness even if you were magically and momentarily at the centre but somehow affected only by the gravitational effects of the things that actually are at the centre.

EDIT2: actually, that's not my argument. My argument is that a convincing argument that the centre of mass of the Earth is steady enough will need to explain this stuff away, rather than skimming over it as this thread has.

if you were momentarily at the centre of mass of the Earth but otherwise unaffected by a large part of what dictates the centre of mass of the Earth, you'd be pulled towards the moon.

No, because you'd share in the orbital motion around the common center (which is inside the earth, BTW). If the earth suddenly disappeared, you'd continue to "orbit" the moon, but the moon would pretty much take off in a straight line, (except for solar gravitation) unconcerned with your paltry hundred kilograms. As the moon receded, you'd also take on a solar orbit.

there's a really big body nearby that affects the fluids on this planet

That's the point of tides, the high tide opposite the moon is there because the water is too far away to share equally in the moon's gravitational pull, but the mechanical connection of the earth in the moon/earth orbit tends to sling it off centrifugally. Vice versa the near side facing the moon. But in the center, those would cancel out, and if it didn't, then the entire orbit would be affected.

“Throughout history, poverty is the normal condition of man. Advances which permit this norm to be exceeded — here and there, now and then — are the work of an extremely small minority, frequently despised, often condemned, and almost always opposed by all right-thinking people. Whenever this tiny minority is kept from creating, or (as sometimes happens) is driven out of a society, the people then slip back into abject poverty. This is known as "bad luck.”

My argument is that a convincing argument that the centre of mass of the Earth is steady enough will need to explain this stuff away, rather than skimming over it as this thread has.

I just calculated the effect of the moon alone on the person at the center of the Earth, it was negligible. I can integrate over that plot gnolam posted (do you by the way know the scale of that plot? The gravity ranges by at most +/- 5 cm/s^2 over the surface of the earth (e.g. see this chart) ) and also arrive at some negligible answer. I won't of course, because it is apparent to me that it is negligible, any of those effects you call attention to would apply equally (or even in greater magnitudes) at the surface of the Earth where each one of us is sitting. Those effects are negligible, since I don't feel them. I won't feel them at the center of the Earth either, where they will be far far smaller.